1. The problem statement, all variables and given/known data How much does the auto - dissociation of water contribute to the pH a 10e+3M aqueous solution of sulphuric acid? 2. Relevant equations Ka = [H30+][SO42-}/[HSO4-] = 1.1e-2 Kw = [OH-][H30+] 3. The attempt at a solution Sulphuric acid is a strong acid in the first hydrolysis step: H2SO4 + H20 -> H30+ + HSO4- Sulphuric acid will dissociate completely to form 10^3M of both H30+ and HSO4- I also neglected the auto dissociation of water, which i will come back to later. Leaving the first hydrolysis step aside, looking at the second hydrolysis step: HSO4- + H20 <=> SO42- + H30+ We know that the initial concentration of HSO4- is 10^3M from the first hydrolysis step. 400*Ka = 4.4 >> 10^3, so the equilibrium concentration of HSO4- is not equal to the initial concentration. so [HSO4-] = initial [HSO4-] - [SO42-]formed the concentration of h30+ formed from the second step is equal to that of so42-: [H3O+] = [S042-] combining the above facts, i got a quadratic equation linking Ka and [H30+] let x = [H30+] x^2 + xKa - Ka[HSO4-] = 0 Solving this gives x = 9.226e-4, so [H30+] formed from second hydrolysis step is 9.226e-4M (the other root is negative and is ignored) As the [H30+] formed from the second hydrolysis step (9.226e-4 M) is quite close to the [H30+] formed from the first step (1.0 e-3M), we cannot assume that the total [H30+] is governed by the first dissociation step only. The total [H30+] is the sum of the [H30+] from both the 1st step and the 2nd step, giving [H30+] = 1.923e-3M To find the effect of the auto dissociation of water, I used the Kw relation. Kw = [OH-][H30+] , where [OH-] and [H30+] refer to the total concentrations present in the solution. Water is the only species that will dissociate and form OH- ions, so the total [OH-] in the solution is equal to the [OH-] from the dissociation of water, which is also equal to the [h30+] formed from the dissociation of water. Knowing this, we find [h30+] (from water) = [OH-] = Kw/[H30+], giving the h30+ contribution of the auto dissociation of water to be 5.2e-12M however the answer is 5.43e-12M. What have I done wrong? Thank you very much in advance.